I'm learning scheme using DrRacket version 7.0.
I have copied, and modified, the following procedure from this SO answer.
#lang racket
(define sort-asc-by-second
(lambda (lst)
(sort lst
(lambda (x y) (< (cdr x) (cdr y))))))
And it works perfectly.
Now I have change the #lang racket instruction with this one #lang r5rs.
And DrRacket complains with the following error:
sort: unbound identifier in: sort
Do I forget to add something to DrRacket (libraries, modules, etc.)? Or maybe sort is not available in R5RS.
According to the R5RS report there is no such thing as sort. You may fetch the reference implementation for sort from SRFI-95 Sorting and merging. It's not 100% compatible with the #lang racket one. R5RS doesn't have libraries so the portable way would be to load it or inline it.
R6RS has superseded R5RS and it has list-sort in its standard library.
R7RS-Large, which is not yet fully ratified, uses SRFI-132 as it sorting library (scheme sort) which also has list-stable-sort which is the same as the R6RS list-sort.
In Racket you can mix and match libraries from all the languages it supports, but it does not work for lists that are implemented differently and it is a lock in. Being dependent on a different languages library means you cannot run the code with other implementations, like Ikarus and Chez.
Related
Can any case of using call/cc be rewritten equivalently without using it?
For example
In (g (call/cc f)), is the purpose of f to evaluate the value of
some expression, so that g can be applied to the value?
Is (g (call/cc f)) always able to be rewritten equivalently
without call/cc e.g. (g expression)?
In ((call/cc f) arg), is the purpose of f to evaluate the
definition of some function g, so that function g can be
applied to the value of arg?
Is ((call/cc f) arg) always able to be rewritten equivalently
without call/cc e.g. (g arg)?
If the answers are yes, why do we need to use call/cc?
I am trying to understand the purpose of using call/cc, by contrasting it to not using it.
The key to the direct answer here is the notion of "Turing equivalence". That is, essentially all of the commonly used programming languages (C, Java, Scheme, Haskell, Lambda Calculus etc. etc.) are equivalent in the sense that for any program in one of these languages, there is a corresponding program in each of the other languages which has the same meaning.
Beyond this, though, some of these equivalences may be "nice" and some may be really horrible. This suggests that we reframe the question: which features can be rewritten in a "nice" way into languages without that feature, and which cannot?
A formal treatment of this comes from Matthias Felleisen, in his 1991 paper "On the Expressive Power of Programming Languages" (https://www.sciencedirect.com/science/article/pii/016764239190036W), which introduces a notion of macro expressibility, pointing out that some features can be rewritten in a local way, and some require global rewrites.
The answer to your original question is obviously yes. Scheme is Turing-complete, with or without call/cc, so even without call/cc, you can still compute anything that is computable.
Why "it is more convenient than writing the equivalent expression using lambda"?
The classic paper On the Expressive Power of Programming Languages by Matthias Felleisen gives one answer to this question. Pretty much, to rewrite a program with call/cc to one without it, you might potentially need to transform your whole program (global transformation). This is to contrast some other constructs that only need a local transformation (i.e., can be written as macro) to remove them.
The key is: If your program is written in continuation passing style, you don't need call/cc. If not, good luck.
I whole-heartedly recommend:
Daniel P. Friedman. "Applications of Continuations: Invited Tutorial". 1988 Principles of Programming Languages (POPL88). January 1988
https://cs.indiana.edu/~dfried/appcont.pdf
If you enjoy reading that paper, then check out:
https://github.com/scheme-live/bibliography/blob/master/page6.md
Of course anything that is written with call/cc can be written without it, because everything in Scheme is ultimately written using lambda. You use call/cc because it is more convenient than writing the equivalent expression using lambda.
There are two senses to this question: an uninteresting one and an interesting one:
The uninteresting one. Is there some computation that you can do with call/cc that you can't do in a language which does not have it?
No, there isn't: call/cc doesn't make a language properly more powerful: it is famously the case that a language with only λ and function application is equivalent to a universal Turing machine, and thus there is no (known...) more powerful computational system.
But that's kind of uninteresting from the point of view of programming-language design: subject to the normal constraints on memory &c, pretty much all programming languages are equivalent to UTMs, but people still prefer to use languages which don't involve punching holes in paper tape if they can.
The interesting one. Is it the case that call/cc makes some desirable features of a programming language easier to express?
The answer to this is yes, it does. I'll just give a couple of examples. Let's say you want to have some kind of non-local exit feature in your language, so some deeply-nested bit of program can just say 'to hell with this I want out', without having to climb back out through some great layer of functions. This is trivial with call/cc: the continuation procedure is the escape procedure. You can wrap it in some syntax if you want it to be nicer:
(define-syntax with-escape
(syntax-rules ()
[(_ (e) form ...)
(call/cc (λ (e) form ...))]))
(with-escape (e)
... code in here, and can call e to escape, and return some values ...)
Can you implement this without call/cc? Well, yes, but not without either relying on some other special construct (say block and return-from in CL), or without turning the language inside out in some way.
And you can build on things like this to implement all sorts of non-local escapes.
Or, well, let's say you want GO TO (the following example is Racket):
(define (test n)
(define m 0)
(define start (call/cc (λ (c) c)))
(printf "here ~A~%" m)
(set! m (+ m 1))
(when (< m n)
(start start)))
Or, with some syntax around this:
(define-syntax-rule (label place)
(define place (call/cc identity)))
(define (go place)
(place place))
(define (horrid n)
(define m 0)
(label start)
(printf "here ~A~%" m)
(set! m (+ m 1))
(when (< m n)
(go start)))
So, OK, this perhaps is not a desirable feature of a programming language. But, well, Scheme doesn't have GO TO right, and yet, here, it does.
So, yes, call/cc (especially when combined with macros) makes a lot of desirable features of a programming language possible to express. Other languages have all these special-purpose, limited hacks, Scheme has this universal thing from which all these special-purpose hacks can be built.
The problem is that call/cc doesn't stop with the good special-purpose hacks: you can also build all the awful horrors that used to blight programming languages out of it. call/cc is like having access to an elder god: it's really convenient if you want dread power, but you'd better be careful what comes with it when you call, because it may well be an unspeakable horror from beyond spacetime.
An easy use of call/cc is as a bail out. eg.
;; (1 2) => (2 4)
;; #f if one element is not a number
(define (double-numbers lst)
(call/cc
(lambda (exit)
(let helper ((lst lst))
(cond ((null? lst) '())
((not (number? (car lst))) (exit #f))
(else (cons (* 2 (car lst)) (helper (cdr lst)))))))))
So to understand this. If we are doing (double-numbers '(1 2 r)) the result is #f, but the helper has done (cons 1 (cons 2 (exit #f)))
Without call/cc we see the continuation would be whatever called double-numbers since it actually return normally from it. Here is an example without call/cc:
;; (1 2) => (2 4)
;; #f if one element is not a number
(define (double-numbers lst)
(define (helper& lst cont)
(cond ((null? lst) (cont '()))
((not (number? (car lst))) #f) ; bail out, not using cont
(else (helper& (cdr lst)
(lambda (result)
(cont (cons (* 2 (car lst)) result)))))))
(helper& lst values)) ; values works as an identity procedure
I imagine it gets harder pretty quick. Eg. my generator implementation. The generator relies on having access to continuations to mix the generator code with where it's used, but without call/cc you'll need to do CPS in both the generator, the generated generator and the code that uses it.
I am a undergraduate who wants to go through "The Scheme programming language" as a self-study.
Here is a simple program and I named it as "reciprocal.ss"
(define reciprocal
(lambda (n)
(if(= n 0)
"oops!"
(/ 1 n))))
Then I wanted to load my procedure:
(load "reciprocal.ss")
It produces this error:
reciprocal.ss:1:0: #%top-interaction: unbound identifier;
also, no #%app syntax transformer is bound in: #%top-interaction
I did each parts as what the book says. Perhaps I am just making a rookie mistake. Any insight would be appreciated.
Since load uses eval, using it outside of a REPL generally will not work — for reasons described in Namespaces
Using racket/load can work for you here however:
loader.ss
#lang racket/load
(load "reciprocal.ss")
(display (reciprocal 10))
reciprocal.ss
(define reciprocal
(lambda (n)
(if (= n 0) "oops!"
(/ 1 n))))
In Racket (and Scheme at large) has a more complex story than the average language regarding running external code. In general, you should use import when you want to directly 'include' a file, you should use provide/require when you want to establish module boundaries and you should use load when you are sophisticated enough to be stretching the limits of either.
The simplest approach is not to use load at all.
In "reciprocal.ss" make the first lines:
#lang racket
(provide (all-defined-out))
(define reciprocal
(lambda (n)
(if (= n 0)
"oops!"
(/ 1 n))))
Then use (require "reciprocal.ss") in the file where you need to use the function reciprocal.
The load mechanism was used back in the good old days before module systems had arrived. Writing (load "foo.ss") basically works as if you manually pasted the contents of foo.ss into the repl and excecuted it. This means that the result of your program is dependent of the order of loading files (if you are using side effects). Module systems handle this (and other things too) much better.
I've been learning Scheme through the Little Schemer book and it strikes me as odd that a file in Scheme / Lisp isn't a list of statements. I mean, everything is supposed to be a list in Lisp, but a file full of statements doesn't look like a list to me. Is it represented as a list underneath? Seems odd that it isn't a list in the file.
For instance...
#lang scheme
(define atom?
(lambda (x)
(and (not (pair? x)) (not (null? x)))))
(define sub1
(lambda (x y)
(- x y)))
(define add1
(lambda (x y)
(+ x y)))
(define zero?
(lambda (x)
(= x 0)))
Each define statement is a list, but there is no list of define statements.
It is not, because there is no practical reasons for it. In fact, series of define statements change internal state of the language. Information about the state can be accessible via functions. For example , you can ask Lisp if some symbol is bound to a function.
There is no practical benefit in traversing all entered forms (for example, define forms). I suppose that this approach (all statements are elements of a list) would lead to code that would be hard to read.
Also, I think it not quite correct to think that "everything is supposed to be a list in Lisp", since there are also some atomic types, which are quite self-sufficient.
When you evaluate a form, if the form defines something, that definition is added to the environment, and that environment is (or can be) a single list. You can build a program without using files, by just typing definitions into the REPL. In Lisp as in any language, the program “lives” in the run-time environment, not the source files.
Is it true that this is an S-expression?
xyz
asks The Little Schemer. but how to test?
syntactically, i get how to test other statements like
> (atom? 'turkey)
and
> (list? '(atom))
not entirely sure how to test this...
> (list? '(atom turkey) or)
as it just returns...
or: bad syntax in: or
but anyway, knowing how to test for S-expressions is foxing me
so, as per usual, any illumination much appreciated
An "S-expression" is built of atoms via several (possibly zero) cons applications:
(define (sexp? expr)
(or
; several cases:
(atom? expr)
; or
(and (pair? expr) ; a pair is built by a cons
(sexp? (car expr)) ; from a "car"
(sexp? .........)) ; and a "cdr"
)))
This is practically in English. Nothing more to say about it (in code, I mean). Except, after defining the missing
(define (atom? x)
(not (pair? x)))
we see that (sexp? ...) can only return #t. This is the whole point to it: in Lisp, everything is an S-expression – either an atom, or a pair of S-expressions.
The previous answer is correct -- Scheme (and Lisp) are languages that are based on S-expressions. And the provided code is a great start.
But it's not quite correct that everything is an S-expression in those languages. In this case you have an expression that isn't syntactically correct, so the language chokes when it tries to read it in. In other words, it's not an S-expression.
I know it's frustrating, and honestly, not that great of an answer, but this is a great lesson in one of the Golden Rules of Computer Programming: Garbage In, Garbage Out. The fat is that you are going to get these kinds of errors simply because it isn't really feasible, when starting out programming, to test for every possible way that something isn't an S-expression without using the language itself.
In Scheme, procedures like +, -, *, / works on different types of numbers, but we don't much see any other generic procedures.
For example, length works only on list so that vector-length and string-length are needed.
I guess it comes from the fact that the language doesn't really offer any mechanism for defining generic procedure (except cond of course) like "type classes" in Haskell or a standardized object system.
Is there an idiomatic scheme way to handle generic procedures that I'm not aware of ?
Keeping in mind that all "different types of numbers" are all scheme numbers (i.e. (number? n) equals #t) - this behavior actually makes sense. +, -, *, / and all other arithmetic operators operate on numbers only (even though in other languages they would be classified as different number types int, long, float, etc...) This is due to the fact that you can't explicitly declare number types in scheme.
If you really need a generic solution, besides using external libraries, the easiest way is to roll your own:
(define org-length length)
(define (length x)
(cond
((string? x) (string-length x))
((vector? x) (vector-length x))
; keep going ...
(else (org-length x))))
No, but you can build your own. Welcome to Scheme!
In the past I've used Swindle to provide generic functions. It's bundled with PLT Scheme. It worked well for me, but it's been a few years. There may be other alternatives out there now.
Read SICP, sections 2.4 and 2.5, which cover the implementation of procedures that can operate on generic data types by means of attaching "tags" to data objects. It's also in lecture 4-B of that MIT video series.
You really want to have an object system for that. You may want to have a look at Tiny CLOS, for instance, which is the de-facto standard object system for Chicken Scheme (see the reference manual), but seems to be available for most Scheme implementations.
Finally, I found out a very neat solution in PLT Scheme :
(require (rename-in scheme [length list-length]))
(define length
(λ (x)
((cond [(list? x) list-length]
[(string? x) string-length]
[(vector? x) vector-length]
[else (error "whatever")]) x)))
(length '(a b c))
(length "abc")
(length #(1 2 3))